//Pow(x, n) -- 递归练习

class Solution {
public:
    double myPow(double x, int n) {
        return n < 0 ? 1.0 / pow(x, -(long long)n) : pow(x , n);
    }
    double pow(double x, int n)
    {
        if(n == 0) return 1.0;
        double tmp = pow(x, n/2);
        return n % 2 == 0 ? tmp * tmp: x * tmp * tmp;
    }
};